Match the column — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

(A) For pure water at $25^\circ\text{C}$, $K_a = K_b = 1.8 \times 10^{-16}\text{ M}$. Thus, $pK_a = pK_b = 16 - \log 1.8 = -\log K_{a, \text{H}_2\text{O}}$. The arithmetic average of identical values yields the same value ($16 - \log 1.8$ and $-\log K_{a, \text{H}_2\text{O}}$). Matches Right B and C. (B) The pH of a salt formed by a weak acid and a weak base where $K_a = K_b$ is exactly neutral at any temperature, corresponding to exactly $pK_w / 2$. Matches Right A. (C) The autoionization of water is an endothermic process, so at 320 K (higher than 298 K), the value of $K_w$ increases, meaning $pK_w < 14$. The pH of pure water remains exactly $pK_w / 2$, which evaluates to strictly < 7.0. Matches Right A and D. Therefore, correct answer is 1-B,C; 2-A; 3-A,D.